Hej. Poproszę o pomoc w zadaniach i rozwiązanie w załączniku. Dziękuje.

[latex]1.\a_{1}=54\a_{3}=2\a_{2}= sqrt{a_{1}*a_{3}} \(a_{2})^{2}=a_{1}*a_{3}\(a_{2})^{2}=54*2=108\a_{2}= sqrt{108} =6 sqrt{3} [/latex] [latex]2.\a_{1}= sqrt{4} \a_{2}=x+2\a_{3}=3x\\a_{2}=frac{a_{1}+a_{3}}{2}\x+2=frac{ sqrt{4}+3x }{2}\2(x+2)= sqrt{4}+3x\2x+4=sqrt{4}+3x\4-sqrt{4}=x [/latex] [latex]3.\K=k_{0}(1+frac{p}{100})^{n}\k_{0}=3000zl\p=4*frac{1}{4}=1\n=4\\K=3000zl(1+frac{1}{100})^{4}=3000zl*(1,01)^{4}=3121,81zl[/latex] Odp: Pan Jan po roku oszczędzania otrzyma 3121,81zl [latex]4.\a_{4}=7=a_{1}+3r\a_{7}=16=a_{1}+6r\\ left { {{7=a_{1}+3r} atop {16=a_{1}+6r}} ight. \left { {{-7=-a_{1}-3r} atop {16=a_{1}+6r}} ight\9=3r\3=r\7=a_{1}+3r\7=a_{1}+9\-2=a_{1}[/latex] [latex]5.\a_{1}=3\a_{2}=8\r=a_{2}-a_{1}=8-3=5\a_{20}=a_{1}+19r=3+95=98\\S_{20}=frac{a_{1}+a_{20}}{2}*20=frac{3+98}{2}*20=1010[/latex]